$C_5{H}_{12}$ is of the form $Cn{H}_{2n+2}$.
Hence it is a saturated alkane. Further, it is given that it has a quaternary carbon. Hence its structure is:

Longest chain = 3 carbons $⇒$ word root = prop

                                                      Primary suffix = ane

                                                  Secondary prefix = 2,2-Dimethyl

IUPAC name is (c) 2,2-Dimethyl propane    

The numbering is according to the lowest of locants, which gives us two options.

In this case, numbering happens according to alphabetical order. So ethyl group will have the lower number.

word root = hex

primary prefix = cyclo

secondary prefix = 1-Ethyl-2-methyl

primary suffix = ane

IUPAC name is

(c)1-Ethyl-2-methylcyclohexane

Word root = pent

Primary suffix = ane

Lowest locant set = 3,3

Secondary prefixes = 3-(chloromethyl)-3-methyl

IUPAC name is 3-(chloromethyl)-3-methylpentane

Decide the parent chain and give the numbering so that more substituted substituent will get the least number.

parent chain

10 carbons $⇒$ word root $⇒$ dec

single bond $⇒$ suffix $⇒$ ane

substituents at $C_5\text{Â}and\text{Â}{C}_2$

Now,

There is one more same substituent on $C_5$

Therefore,

   5,5-Bis-(1,1-dimethyl propyl)

Bis$⇒$ Because two same substituents are there, and a methyl group at $C_2⇒$ 2-methyl

Therefore,

   5,5-Bis-(1,1-dimethylpropyl)-2-methyldecane

Decide parent chain.

Numbering so that more substituted substituent will get the lowest number.

Five carbons $⇒$ word root = pent

                                    single bonds = ane(suffix)

$→$ substituent at $C_2\text{Â}and\text{Â}{C}_3$

$⇒$ 3-ethyl-2, 3-dimethylpentane(alphabetically)

Therefore,

                (b) 3-Ethyl-2,3-dimethylpentane

Our first step should be to choose the longest chain from the given alkane.
Although there are 9 carbon atoms in total, the longest chain contains only 7 carbon atoms.

Primary suffix = ane

Word root = hept 

If we number the chain, as shown above, the two methyl substituents are at $C_4$ and $C_6$

Now, let's look at another case.

Now we have substituents at $C_2\text{Â}and\text{Â}{C}_4$. In both cases, there are exactly two substituents.
We choose the numbering direction that gives the lowest set of locants.

Hence, secondary prefix = 2,4-dimethyl (since there are two methyl groups-one each at $C_2\text{Â}and\text{Â}{C}_4$

Hence, the IUPAC name is 2,4-Dimethylheptane

 

First, check the parent chain.
Now, give numbers such that the more substituted substituent will get the lowest number.

Parent chain
7 carbon $⇒$ hept (word root)
Single bonds $⇒$ ane (suffix
$→$ substituent at $C_3\text{Â}and\text{Â}{C}_4$ 3-ethyl-4,4-dimethyl(alphabetically)
Therefore, 3-Ethyl-4,4-dimethylheptane

The given compound tertiary butyl chloride is

   
It is numbered as shown.

The longest chain clearly includes 3 carbon atoms

$⇒$  root word = prop

primary suffix = ane

Secondary prefix = 2-chloro-2-methyl

Hence the name (systematic IUPAC) of given compound is:

(d) 2-Chloro-2-methylpropane

$\stackrel{1}{\underset{5}{C{H}_3}}-\stackrel{2}{\underset{4}{CH}}=\stackrel{3}{\underset{3}{CH}}-\stackrel{4}{\underset{2}{C}}\equiv \stackrel{5}{\underset{1}{CH}}$
From lowest sum rule, we will select $1,3$ and not $2,4$

Five carbons$\Rightarrow$ word root = pent

3-Penten-1-yne

Parent chain is
$\stackrel{1}{C}{H}_3-\stackrel{2}{C}{H}_2-\stackrel{3}{\underset{C{H}_3}{\underset{|}{C}}}=\stackrel{4}{C}H-\stackrel{5}{C}{H}_2-\stackrel{C{H}_3}{\stackrel{6|}{C}H}-\stackrel{7}{C}{H}_2-\stackrel{8}{C}{H}_3$

Word root$\Rightarrow$ oct

Double bond$\Rightarrow$suffix$\Rightarrow$ene

Two methyl groups at $C_3$and$C_5\Rightarrow$3,6-dimethyl

Therefore, 3,6-Dimethyl-3-octene